I dont know but i just went on here to look up my answers but it 2
After your first exposure to a pathogen, you have memory T cells that will remember the antigen of the pathogen son in the future if you would come in contact with the same pathogen your body will recognize it right away and be able to kill it much faster
Answer:
7.61 liters
Explanation:
From the question,
Applying Charles law formula,
V₁/T₁ = V₂/T₂...................... Equation 1
Where P₁ = initial volume of of air, V₂ = Final volume of air, T₁ = Initial Temperature of air, T₂ = Final Temperature of air.
make V₂ the subject of the equation
V₂ = (V₁×T₂)/T₁............... Equation 2
Given: V₁ = 5 liters, T₁ -35 °C = (-35+273) K = 238 K, T₂ = 89°C = (89+273) K = 362 K
Substitute these values into equation 2
V₂ = (5×362)238
V₂ = 1810/238
V₂ = 7.61 liters
Answer:
A. The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
Explanation:
- The amount of heat absorbed by water (Q) can be calculated from the relation:
<em>Q = m.c.ΔT.</em>
where, Q is the amount of heat absorbed by water,
m is the mass of water,
c is the specific heat capacity of water (c = 4.186 J/g °C),
ΔT is the temperature difference (final T - initial T).
- <u><em>For trial 1:</em></u>
m = 30.0 g, c = 4.18 J/g °C, ΔT = 40.0 °C – 0.0 °C = 40.0 °C
<em>∴ Q = m.c.ΔT</em> = (30.0 g)(4.18 J/g °C)(40.0 °C) = <em>5023 J.</em>
- <u><em>For trial 2:</em></u>
m = 40.0 g, c = 4.18 J/g °C, ΔT = 40.0 °C – 10.0 °C = 30.0 °C
<em>∴ Q = m.c.ΔT</em> = (40.0 g)(4.18 J/g °C)(30.0 °C) = <em>5023 J.</em>
<em>A. The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both. </em>
